Learn Coulomb's law gives the electrostatic interaction force between the two point charges. Practice coulomb's law problems and representation of coulomb's law in vector form.
\(F_{12} = F_{21} = \dfrac{K|q_1||q_2|}{r^2}\)
where \(K\)= electrostatic constant = \(\dfrac{1}{4\pi\epsilon_0}=9× 10^9\,Nm^2/C^2\)
\(\epsilon_o = \) Permittivity of free space = \(8.85 × 10^{-12} C^2/ Nm^2\)
Note: Coulomb's law is only applicable for charges at rest.
A Doubled
B Halved
C One - fourth
D Tripled
Since charge q_{1} repels q_{2}, Force \(\vec F_{21}\) acts towards right
\(\vec F _{21} = \dfrac{K|q_2||q_1|}{(r_1)^2}\) towards right
Since charge q_{3} repels q_{2 }, Force \(\vec F_{23}\) acts towards left
\(\vec F _{23} = \dfrac{K|q_2||q_3|}{(r_2)^2}\) towards left
\(\vec F_2 =\vec F_{21} +\vec F_{23} \)
\(\vec F_2 = \dfrac{K q_1q_2}{r_1^2}\) towards right + \( \dfrac{K q_2q_3}{r_2^2}\) towards left
A 109 N towards right
B 1010 N towards left
C 0.1 N towards right
D 1012 N towards left
Position vector of charge \( q_2 = (x_2 \, \hat i + y_2\,\hat j)\)
Position vector of charge \( q_3 = (x_3 \, \hat i + y_3\,\hat j)\)
\(\vec F_{31} = \dfrac{Kq_1q_3}{|\vec r_{31}|^2} \hat r_{31}\)
where position vector \(\vec r_{31} = \vec r_3 -\vec r_1 \)
\(\vec F_{32} = \dfrac{Kq_2q_3}{|\vec r_{32}|^2} \hat r_{32}\)
where position vector \(\vec r_{32} = \vec r_3 -\vec r_2 \)
\(\vec F _3 = \vec F_{31} + \vec F_{32}\)
\(\vec F_{3} = \dfrac{Kq_1q_3}{|\vec r_{31}|^2} \hat r_{31} + \dfrac{Kq_2q_3}{|\vec r_{32}|^2} \hat r_{32}\)
A \(504 (-\hat i - \hat j)\ N\)
B \(270 (-\hat i + \hat j)\ N\)
C \(140 (-\hat i - \hat j)\ N\)
D \(120 (\hat i +\hat j)\ N\)
A dipole is a combination of two equal and opposite charges, separated by a small distance d.
To calculate force on a charge placed at a point on equatorial line of dipole, consider a dipole and a point P on equatorial line of dipole at a distance r.
\(\vec F_{+q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(\sqrt {r^2 +\ell^2)}^2} \) (B to P)
\(\vec F_{+q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(r^2 +\ell^2)} \) (B to P)
\(\vec F_{-q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(\sqrt {r^2 +\ell^2)}^2} \) (P to A)
\(\vec F_{-q} = \dfrac{1}{4\pi\epsilon_o} \dfrac{q.q_0}{(r^2 +\ell^2)} \) (P to A)
Here , \(|\vec F_{+q}| =|\vec F_{-q}| = F \)
Net force = \(2 × \dfrac{1}{4\pi\epsilon_o} × \dfrac{q.q_0}{(r^2+\ell^2)} cos\theta\)
Here , \(cos \theta = \dfrac{\ell}{\sqrt{r^2+\ell^2}}\)
Net force = \(2 × \dfrac{1}{4\pi\epsilon_o} × \dfrac{q.q_0}{(r^2+\ell^2)} . \dfrac{\ell}{\sqrt{r^2+\ell^2}}\)
Net force = \( \dfrac{2}{4\pi\epsilon_o} × \dfrac{q.q_0\,\ell}{(r^2+\ell^2)^{3/2}}\)
\(\vec F_{net} = \dfrac{2}{4\pi\epsilon_o} × \dfrac{q.q_0\,\ell}{(r^2+\ell^2)^{3/2}} (-\hat i)\)
A \(9\sqrt2 × 10^3\ \hat i\,N \)
B \(6× 10^8\ \hat i\,N\)
C \(4× 10^3\ \hat i\,N\)
D \(2× 10^6\ \hat i\,N\)
\(|\vec F_{12}| = |\vec F_{21}|= \dfrac{K |q_1||q_2|}{ r^2} \)
here, \(|q_1|,|q_2|\) are magnitude of both the charges
Case 1
Consider two point charges, q_{1} and q_{2,} placed at distance 'r' apart.
Force on charge q_{1} due to q_{2} \((\vec F_{12})\) in vector form
\(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)
here, q_{1 }and q_{2} are the magnitude of charge along with sign
\(\vec r_{12}\) is the position vector of \(q_1\) with respect to \(q_2 = \vec r_1 - \vec r_2 \)
where, \(\vec r_1=\)position vector of q_{1}
\(\vec r_2=\)position vector of q_{2}
\(\hat r_{12}\) is the unit vector directed towards q_{1} from q_{2}
Conclusion - Hence, it is concluded that the direction of force \(\vec F_{12}\) is along \(\hat r_{12}\).
Force on charge q_{2} due to q_{1} \((\vec F_{21})\) in vector form
\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)
here, q_{1} and q_{2} are the magnitude of charge along with sign
\(\vec r_{21}\) is the position vector of \(q_2\) with respect to \(q_1 = \vec r_2 - \vec r_1 \)
where, \(\vec r_1=\)position vector of q_{1}
\(\vec r_2=\)position vector of q_{2}
\(\hat r_{21}\) is the unit vector directed towards q_{2} from q_{1}
Conclusion - Hence, it is concluded that the direction of force \(\vec F_{21}\) is along \(\hat r_{21}\).
Case 2
Consider two charges, q_{1} and – q_{2 ,} placed at 'r' distance apart.
Force on charge q_{1} due to – q_{2} \((\vec F_{12})\) in vector form
\(\vec F_{12} = \dfrac{K q_1(-q_2)}{|\vec r_{12}|^2} \hat r_{12}\)
\(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} (-\hat r_{12})\)
Conclusion - From above formula it is proved that direction of force is along \(- \hat r_{12}\).
Since, q_{1} and q_{2} are opposite in nature so, q_{1} will attract – q_{2} . So, force \((\vec F_{21})\) will act towards q_{1}.
\(\vec F_{21} = \dfrac{K q_1(-q_2)}{|\vec r_{21}|^2} \hat r_{21}\)
\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} (-\hat r_{21})\)
Conclusion - From above formula it is concluded that direction of force is along \(- \hat r_{21}\) .
Case 3
Consider two charges, –q_{1} and –q_{2 , }placed at 'r' distance apart.
Since charge are of same nature so, –q_{2} repels –q_{1} and force \(\vec F_{12}\) acts away form –q_{2}.
\(\vec F_{12} = \dfrac{K (-q_1)(-q_2)}{|\vec r_{12}|^2} \hat r_{12}\)
\(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)
Conclusion - Hence, It is concluded that the direction of force \(\vec F_{12}\) is along \( \hat r_{12}\) .
Since, charge are same nature so, –q_{1} will repel – q_{2} and force \(\vec F_{21} \) will act away form - q_{1}.
\(\vec F_{21} = \dfrac{K (-q_1)(-q_2)}{|\vec r_{21}|^2} \hat r_{21}\)
\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)
Conclusion - Hence , it is concluded that direction of force \(\vec F_{21}\) is along \(\hat r_{21}\) .
Note - In vector form of Coulomb's law we don't need to care about the direction of force, just put the values of q_{1} and q_{2} along with sign, vector will take care of direction .
A along \(\hat r_{21}\)
B along \(-\hat r_{21}\)
C along \(-\hat r_{12}\)
D along \(\hat r_{12}\)
\(\vec F_{12} = \dfrac{K q_1q_2}{|\vec r_{12}|^2} \hat r_{12}\)
here , q_{1 }and q_{2} are the magnitude of charge along with sign
\(\vec r_{12}\) is the position vector of \(q_1\) with respect to \(q_2 = \vec r_1 - \vec r_2 \)
where \(\vec r_1=\)position vector of q_{1}
\(\vec r_2=\)position vector of q_{2}
\(\hat r_{12}\) is the unit vector directed towards q_{1} from q_{2}.
Conclusion - Hence, it is concluded that the direction of force \(\vec F_{12}\) is along \(\hat r_{12}\).
\(\vec F_{21} = \dfrac{K q_1q_2}{|\vec r_{21}|^2} \hat r_{21}\)
here, q_{1} and q_{2} are the magnitude of charge along with sign
\(\hat r_{21}\) is the position vector of \(q_2\) with respect to \(q_1 = \vec r_2 - \vec r_1 \)
where \(\vec r_1=\)position vector of q_{1}
\(\vec r_2=\)position vector of q_{2}
\(\hat r_{21}\) is the unit vector directed towards q_{2} from q_{1}.
Conclusion - Hence, it is concluded that the direction of force \(\vec F_{21}\) is along \(\hat r_{21}\).
A \(9× 10^{–3} \,N (-\hat j)\)
B \(9× 10^6 \,N (-\hat i)\)
C \(9× 10^4\, N (\hat j)\)
D \(18× 10^9\, N (-\hat j)\)